# Basic Math Question (matrices)

1. Nov 19, 2008

### natural

1. The problem statement, all variables and given/known data
Show that the matrix B= 4 -2 -2
1 1 -1
2 -2 0
B^2-3B+2I=0

Hence show that B^3=7B-6I

3. The attempt at a solution
Now i have already worked out the quadratic equation and the matirx B satisfies the equaition B^2-3B+2I=0.

My problem is i cant seem to prove the next part i am stuck here is how for i got

B^3=7B-6I
multiplying by B^2 gives

B^4 - 3B^3 + 2B^2I = O

multiply b B^-1 (B inverse) gives

IB^3 - 3IB^2 = 2IB =0

this is where i am stuck.. thanks for any help given...question dued tomorrow.

2. Nov 19, 2008

### Staff: Mentor

You said you showed that B satisfies B^2 - 3B + 2I = 0. Now you want to show that the same matrix satisfies B^3 - 7B + 6I = 0.

These matrix equations work much like the equations you worked with earlier in your math classes with regard to factorization. For example, your first equation factors into
(B - 2I)(B - I) = 0. Instead of doing what you did before (multiplying B by itself, subtracting 3*B, and then adding 2*I), you could have calculation B - 2I and B - I and multiplied these two matrices together.

As it turns out, your second equation can also be factored, and one of the factors is B^2 - 3B + 2I.

3. Nov 19, 2008

### HallsofIvy

Staff Emeritus
You are given B so just multiply to get B3. Then calculuate 7B- 6I and see if they are the same!

4. Nov 19, 2008

### natural

yes i agree but we cant do it that we we have to use the quadratic equation to prove that B^3 = 7B-6I

5. Nov 19, 2008

### HallsofIvy

Staff Emeritus
Then do the obvious thing: divide x3- 7x+ 6 by x2- 3x+ 2.

6. Nov 19, 2008

### cloudyhill

B^2=3b-2i
b^3=3b^2-2b=3(3b-2i)-2b=7b-6i

7. Nov 19, 2008

### natural

and how will that help me i get x+3 :|

8. Nov 19, 2008

### natural

Thanks alot cloudyhill that helps greatly!

9. Nov 19, 2008

### gabbagabbahey

That tells you that $B^3-7B+6I=(B+3I)(B^2-3B+2I)$ but you already showed that $B^2-3B+2I=0$, sooo $B^3-7B+6I=\ldots$?